(This paper is adapted from a post I made to the Math Message Board)

 

You can easily see both the "little dot" inner product and the outer product here...

If we take the two complex numbers depicted in the graph above,  and , and multiply on the left by , we get the mapping of these two complex numbers to vectors I mentioned in my previous post.  That is, the vectors representing the complex numbers are  and , as depicted above.

From the two complex numbers here, their inner and outer product is:

 

 

 

 

which can be easily seen from the graph above.

The inner product of two vectors is the scalar, .  Remember that the cosine equals zero at 90 degrees... so, the law of cosines just reduces to the Pythagorean theorem when theta here equals zero.  From this, and the graph above, it can be seen that the inner product is a proportional measure of two vectors' deviance from being perpendicular to each other... and this is the proportional relation necessary to apply a correction to the Pythagorean theorem when the angle between two vectors isn't a right angle.

The outer product, or "wedge" product, is equal to the directed area of the parallelogram depicted above, which is the yellow region on the graph.  "Directed" because the sign of the outer product changes depending on which direction you take the two vectors, a and b, in the graph above.  If you go along the bottom of the parallelogram first, and then along the right side, this is .  If you go along the left side of the parallelogram first, and then across its top, this is .  Taking one, or the other, of these paths around two sides of the parallelogram changes the sign of the "directed" area, which is a bivector.  That is, .  As indicated here and from the graph above, the outer product of two vectors is a directed area, or a bivector, where .

Now, you can see from the graph above, that the vector c is the vector sum of the vectors a and b.  So, we start here, square both sides with the inner product, and see what develops...

 

Since we know from above that , it's evident that what we have here is a vector equation that upon changing to a, b, and c to their scalar counterparts corresponding to the sides of the triangle depicted in the graph above, is, essentially the law of cosines.

There is, of course, some apparent variance in sign.  That is, the law of cosines is:

 

First of all, I believe an assumption is made that a, b, and c are positive scalars.  Following from this, the negation is justified by the fact that the angle opposite the hypotenuse, that is, opposite to the side represented by vector c in the graph above, is not the angle used in the inner product.  Rather, it's the other angle of the parallelogram, an angle supplementary to the angle used to take the inner product... that is, it's the angle equal to .  Recalling that the cosine is positive in the 1^st quadrant and negative in the 2^nd quadrant, and that the reference angles for supplementary angles are equal, it follows that the cosine of the supplementary angle we need for our triangle is the negative of the angle taken in the inner product.

Since I believe it's evident from the graph above (I chose easy numbers), and using the Pythagorean theorem, that a = 7 and b = 5.  Since we know the inner product of the two vectors, a and b, to be 21...

 

 

Which should agree with the familiar formula from trigonometry using scalars.  Since I chose such easy numbers, it's evident from the relations of a right triangle that the cosine of the angle theta above is equal to , which implies that the cosine of it's supplementary angle, the angle we need for our triangle, is .  Consequently,

 

I believe the same line of reasoning, utilizing the inner product again, will yield the formulas for the other sides, opposite the vectors a and b, with only minor adjustments.

 

 

References:

 

New Foundations for Classical Mechanics, 2^nd Edition, by David Hestenes,  1999 Kluwer Academic Publishers.  Particularly section 1-4, “The Inner Product”, pp. 16-20.